On Noetherian ness as modules over base ring, of ideals, of an algebra

Question

Let $R$ be a commutative ring with unity and $A$ be a commutative unital $R$-algebra.

(i) Let $I$ be an ideal of $A$ and $a\in A$ be such that $I+Aa$ and $(I:a)=\{ x\in A : ax \in I\}$ are Noetherian $R$-modules. Then is $I$ Noetherian as an $R$-module ?

(ii) Let $I$ be an ideal of $A$ and $a,b\in A$ be such that $ab\in I$ and $I+Aa$ and $I+Ab$ are finitely generated as $R$-modules. Then is $I$ Noetherian as an $R$-module ?

Answer 1

Hints for (i)

• We have a surjective $A$-linear map $\;\begin{aligned}[t](I:a)&\longrightarrow I\cap Aa\\x&\longmapsto xa\end{aligned}$
• Consider the short exact sequence: $$0\longrightarrow I\cap Aa\longrightarrow I\oplus Aa\longrightarrow I+ Aa\longrightarrow 0.$$